Sum of Dilogarithm and its Complex Conjugate $$\newcommand{\dl}[1]{\operatorname{Li}_2\left( #1\right)}$$
I have the following expression involving Dilogarithims, where $z\in\mathbb{R}$ and $b\in\mathbb{C}\setminus\lbrace 0\rbrace$:
$$f(z,b) = \dl{\frac{z}{b}}-\dl{-\frac{z}{b}}+\dl{\frac{z}{b^*}}-\dl{-\frac{z}{b^*}}$$
In the above, I mean that $b^*$ is the complex conjugate. I have this result from a definite integral, and would like to evaluate $h(b)=\lim_{z\to\infty} f(z,b)$, giving me a result that only depends on $b$. In the above definition of $f$, each term individually diverges when I take the limit. I'm certain that $\lim_{z\to\infty}f(z,b)$ is convergent though, so I'd like to find some way of simplifying $f$ to allow me to take the limit. I'm hoping someone can point me to a relation involving $\dl{z}$, such as $$\dl{z}-\dl{-z}$$ or $$\dl{\frac{z}{b}}+\dl{\frac{z}{b^*}}$$ I am certain one of those relations will help me, but I couldn't find anything like that.
Does someone maybe have an idea, how to evaluate this?
Thanks in advance!
Regards
Zyrax
 A: $$\newcommand{\dl}[1]{\operatorname{Li}_2\left( #1\right)}$$
From the definition $\;\displaystyle\dl z:=\sum_{n=1}^\infty \frac {z^n}{n^2}\;$ we have of course :
\begin{align}
\tag{1}\dl{\overline{z}}&=\overline{\dl{z}},\quad z\in\mathbb{C}\\
&\text{and}\\
\tag{2}\dl{-z}&=-\dl{z}+\frac 12 \dl{z^2}\\
&\text{since}\\
\dl{-z}&=\sum_{n=1}^\infty \frac {(-z)^n}{n^2}=-\sum_{n=1}^\infty \frac {z^n}{n^2}+2\sum_{k=1}^\infty \frac {z^{2k}}{(2k)^2}\\
&\text{but also}\\
\tag{3}\dl{z}-\dl{-z}&=\sum_{n=1}^\infty \frac {z^n-(-z)^n}{n^2}=2\sum_{k=0}^\infty \frac {z^{2k+1}}{(2k+1)^2},\quad |z|<1\\
\end{align}
This allows to rewrite (for $\,x\in\mathbb{R}\,$ and $\,z:=\dfrac xb$) :
\begin{align}
f(x,b) &= \dl{\frac{x}{b}}-\dl{-\frac{x}{b}}+\dl{\frac{x}{\overline{b}}}-\dl{-\frac{x}{\overline{b}}}\\
&= \dl{\frac{x}{b}}-\dl{-\frac{x}{b}}+\overline{\dl{\frac{x}{b}}}-\overline{\dl{-\frac{x}{b}}}\\
\tag{4}&= 2\;\Re\left(\dl{\frac{x}{b}}-\dl{-\frac{x}{b}}\right)\\
&= 4\;\Re\left(\sum_{k=0}^\infty \frac {\left(\frac{x}{b}\right)^{2k+1}}{(2k+1)^2}\right),\quad \text{for $\ \left|\frac{x}{b}\right|<1$ using $(3)$}\\
\tag{5}&=\Re\left(\frac{x}{b}\;\Phi\left(\left(\frac{x}{b}\right)^2, 2, \frac 12\right)\right),\quad\text{(without bounds on $|z|$)}\\
\tag{6}&=4\;\Re\left(\chi_2\left(\frac xb\right)\right)
\end{align}
i.e. as an analytic continuation of the Lerch transcendent
$\;\displaystyle\Phi(z, s, \alpha) := \sum_{n=0}^\infty
\frac { z^n} {(n+\alpha)^s}$
or of the Legendre chi function $\;\displaystyle\chi_s(z) := \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^s}$.
Concerning the asymptotic expression as $x\to +\infty$ the Wikipedia link contains the interesting 
$$\tag{7}\chi_2(x) + \chi_2(1/x)= \frac{\pi}2\left(\frac{\pi}{2}-i|\ln x|\right),\quad x>0$$
The $x>0$ restriction appears not really required and we have in fact for $z\neq 0$ :
$$\tag{8}\chi_2(z) + \chi_2(1/z)= \frac{\pi}2\left(\frac{\pi}{2}+\epsilon\;(\arg(z)-i\ln|z|)\right)\,\quad\text{with}\ \ \epsilon:=\begin{cases}
-1 & \arg(z)>0\\
+1 & \arg(z)\le 0
\end{cases}$$
Since $\;\chi_2(1/z) \to 0\;$ as $\;|z|\to +\infty\;$ with $\,z=\dfrac xb\;$ we obtain simply using $(6)$ and $\arg \dfrac xb=-\arg b$
("arg" is the complex argument i.e. $\;\arg b=\arctan\dfrac {\Im(b)}{\Re(b)}$ if $\Re(b)>0$) :
$$\tag{9}f(x,b)\sim 2\,\pi\,\left(\frac {\pi}2-|\arg(b)|\right),\quad x\to \infty$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{f}\pars{z,b} & \equiv \Li{2}\pars{z \over b} -
\Li{2}\pars{-\,{z \over b}} + \Li{2}\pars{z \over b^{*}} -
\Li{2}\pars{-\,{z \over b^{*}}}
\end{align}

The following approach is valid whenever
$\ds{\pm\,{z \over b}, \pm\,{z \over b^{*}} \not\in
\,\left[\, 0,\infty\, \right)}$. In such a case we can use the identity:
$$
\Li{2}\pars{\xi} =
-\Li{2}\pars{1 \over \xi} - \,{\pi^{2} \over 6} - \half\,\ln^{2}\pars{-\xi}
\,,\qquad\xi \not\in \,\left[\, 0,\infty\, \right)
$$
and $\ds{\ln}$-function branch-cut is the principal one ( along the 'negative $\ds{x}$-axis' ).
$\ds{\lim_{\verts{z} \to \infty}\,\mrm{f}\pars{z,b}}$ is reduced to
\begin{align}
\lim_{\verts{z} \to \infty}\,\,\mrm{f}\pars{z,b} & =
\half\,\lim_{\verts{z} \to \infty}\bracks{\ln^{2}\pars{z \over b} -
\ln^{2}\pars{-\,{z \over b}} + \ln^{2}\pars{z \over b^{*}} -
\ln^{2}\pars{-\,{z \over b^{*}}}}
\end{align}
Other situations
$\ds{\pars{~\mbox{besides}\ \xi \not\in \,\left[\, 0,\infty\, \right)~}}$ can be
explored by means of identities in this link.
